Question

the digit at the tens place of a 2-digit number is three times the digit at the ones place. if the number and the number formed by reversing the digits is 88,find the number

Answer 1

Let the digit at the ones place be x
Then the digit at the tens place will be = 3x
Therefore the number is 10(3x) +x
= 30x+x
= 31x
By reversing the digits we get 10(x)+3x 
= 10x+3x
= 13x
Their sum = 88
⇒31x + 13x = 88
⇒ 44x = 88
⇒ x = 88/44
∴ x = 2
Then the digits are x, 3x = 2 and 3(2)
= 2 and 6
The number = 31x 
= 31(2)
= 62
∴ The number is 62

Answer 2

let the digit in one's place be x.
the digit in ten's place= 3x
the original number= 10(3x) + x  (we multiplied by 10 because we have to find the value in ten's place.)
when the digits are interchanged
the number = 10x + 3x
a/q   10x+3x + (10(3x)+x) = 88
       = 13x+30x+x = 88
       = 44x = 88
       = x =  88/44
       ∴ x= 2
digit in unit's place= 2
digit in ten's place = 3x= 3*2= 6
the original number= 6*10 + 2
                               = 60+2
                               = 62
so the required number is 62.